The idea behind the area trapped between two curves is simpler than it looks.
Subtract the lower curve from the upper one and integrate the difference: that is the area between them.
So first find where the curves meet to set the ends of the interval, then check which curve is on top there.
In the picture, moving the right end b fills the area between y = x and y = x² up to 1/6 at b = 1.
Upper curve minus lower curve
seegongsik.com
The region enclosed by y = x and y = x², from 0 to b; its area S is written at the top
1
👀 What to read in the picture
①The area between two curves is the integral of (upper curve − lower curve)
②On 0 ≤ x ≤ 1, y = x is above and y = x² below
③Moving b to 1 brings the area to 1/6 ≈ 0.1667
④For x > 1, y = x² rises above, and the order flips
The area formula
Area between two curves
S = ∫ab (f(x) − g(x)) dx (f ≥ g)
a and b are the x-coordinates where the curves meet; f is the upper curve and g the lower one
Find the intersections first
The interval comes from the intersections
f(x) = g(x)
Solving the two equations together gives the values of x that become the ends a and b
Worked examples
Example 1
Find the area enclosed by y=x and y=x².
1
Set the two expressions equal to find where they meet.
x = x2 ⇒ x = 0, 1
2
On 0 ≤ x ≤ 1, y = x is on top, so integrate x − x².
S = ∫01 (x − x2) dx = [x22 − x33]01 = 16
▸ 1/6
Trying x = 1/2 gives x = 0.5, more than x² = 0.25, so y = x is on top. Testing one point keeps the subtraction in the right order.
Example 2
Find the area enclosed by y=x² and the line y=2x.
1
Set the two expressions equal to find where they meet.
x2 = 2x ⇒ x = 0, 2
2
On 0 ≤ x ≤ 2, the line y = 2x is on top, so integrate 2x − x².
S = ∫02 (2x − x2) dx = [x2 − x33]02 = 43
▸ 4/3
Between a curve and a line the rule is the same: integrate upper minus lower.
Summary
Strategy
S = ∫ab (upper − lower) dx, a·b are intersections
Fix the interval from the intersections, then integrate upper minus lower
2022 provincial mock exam Math type, adapted
What is the area enclosed by y=x²−2x and the x-axis?
①2/3
②1
③4/3
④2
⑤8/3
▸ ③ 4/3
1
Find where the curve meets the x-axis (y = 0).
x2 − 2x = 0 ⇒ x = 0, 2
2
On 0 ≤ x ≤ 2 the curve lies below the x-axis, so integrate 0 − (x² − 2x).
S = ∫02 (2x − x2) dx = 43
🎯 Exam Points
①Area means integrating (upper − lower)
②Take the ends of the interval at the intersections
③If the curve is below the x-axis, integrate (0 − curve)
④If the order changes partway, split the interval there
⑤Even when writing it with an absolute value, check the sign on each piece when calculating